由于我们无法在 MySQL 中使用 MINUS 查询,因此我们将使用 LEFT JOIN 来模拟 MINUS 查询。可以借助以下示例来理解:
示例
在此示例中,我们有两个表,即 Student_detail 和 Student_info,其内容如下数据 –
mysql> Select * from Student_detail; +-----------+---------+------------+------------+ | studentid | Name | Address | Subject | +-----------+---------+------------+------------+ | 101 | YashPal | Amritsar | History | | 105 | Gaurav | Chandigarh | Literature | | 130 | Ram | Jhansi | Computers | | 132 | Shyam | Chandigarh | Economics | | 133 | Mohan | Delhi | Computers | | 150 | Rajesh | Jaipur | Yoga | | 160 | Pradeep | Kochi | Hindi | +-----------+---------+------------+------------+ 7 rows in set (0.00 sec) mysql> Select * from Student_info; +-----------+-----------+------------+-------------+ | studentid | Name | Address | Subject | +-----------+-----------+------------+-------------+ | 101 | YashPal | Amritsar | History | | 105 | Gaurav | Chandigarh | Literature | | 130 | Ram | Jhansi | Computers | | 132 | Shyam | Chandigarh | Economics | | 133 | Mohan | Delhi | Computers | | 165 | Abhimanyu | Calcutta | Electronics | +-----------+-----------+------------+-------------+ 6 rows in set (0.00 sec)
现在,以下使用 LEFT JOIN 的查询将模拟 MINUS 以返回 Student_info 中的“studentid”值,但不返回 Student_detail 表中的值。
mysql> SELECT studentid from student_info LEFT JOIN Student_detail USING(studentid) WHERE student_detail.studentid IS NULL; +-----------+ | studentid | +-----------+ | 165 | +-----------+ 1 row in set (0.07 sec)
现在,以下查询将为我们提供与上述查询相反的结果,即它将返回 Student_detail 中的“studentid”值,但不会返回 Student_info 表中的值。
mysql> SELECT studentid from student_detail LEFT JOIN Student_info USING(studentid) WHERE student_info.studentid IS NULL; +-----------+ | studentid | +-----------+ | 150 | | 160 | +-----------+ 2 rows in set (0.00 sec)原文来自:www.php.cn
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